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Basics of Electrical Circuits Dr. Sameir Abd Alkhalik Aziez University of Technology Department of Electromechanical Engineering Resources :- Introductory circuit Analysis; by Robert L. Boylestad . Basic Electrical Engineering science; by Mckenzie smith and K.T. Hosie Electrical Technology. د ترجمة، لكھربائية ا الھندسة علم. زكيحمد م& د. أنور مظفرDefinitions :- Electric charge : Fundamental property of sub-atomic particles . Positive charge proton negative charge electron no charge neutron Like charges repel & opposite charges attract. Unit of charge coulomb Symbol and Units:- Difference between a symbol and unit. . ) ( 180 ) ( Unit C Symbol = Η Current :- The movement of electrons is the current which results in work begin done in an electric circuit. Hence, the Electric Current is the rate of flow of charge. We have a current when there is a few of charges; S C A t q I = Δ Δ = The direction of motion of positive charges is opposite to direction of motion of negative charges. Amper ( ) I t q t Sec Coul = Δ Δ = Δ × × = = 6 19 10 10 6 . 1 . . q = Charge that flows through section in time Δ t. Δ Uniform flow of charges direct current ( dc ) - 1 -
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Electrical Circuits - Introduction

Nov 25, 2015

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An introduction to electrical circuits with solved examples for the understanding of basic concepts of circuits.
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  • Basics of Electrical Circuits

    Dr. Sameir Abd Alkhalik AziezUniversity of TechnologyDepartment of Electromechanical Engineering

    Resources :-

    Introductory circuit Analysis; by Robert L. Boylestad . Basic Electrical Engineering science; by Mckenzie smith and K.T. Hosie Electrical Technology. . & .

    Definitions :-

    Electric charge : Fundamental property of sub-atomic particles .

    Positive charge proton

    negative charge electron

    no charge neutron

    Like charges repel & opposite charges attract.

    Unit of charge coulomb

    Symbol and Units:- Difference between a symbol and unit.

    . )(180)( UnitCSymbol =Current :- The movement of electrons is the current which results in work

    begin done in an electric circuit. Hence, the Electric Current is the rate of flow

    of charge.

    We have a current when there is a few of charges;

    SCA

    tqI =

    =

    The direction of motion of positive charges is opposite to direction of motion of

    negative charges.

    Amper ( ) Itq

    tSecCoul =

    ===

    619 10106.1..

    q = Charge that flows through section in time t. Uniform flow of charges direct current ( dc )

    -1-

  • Example: In a copper wire a flow of charge is 0.12 C in a time of 58 ms . Find

    the current in this wire ?

    Solution: AtQ

    tq

    I 06.21058

    12.03===

    =

    Work & Energy:

    dFw = .Where:

    w d: Is the work done by a force ( F ) acting over a displacement ( ). As charges move, subject to various forces they may gain or lose energy.

    = work done = charge in energy wExternal force work done on charges charges gain energy.

    ( Like Sources )

    Work done by charges charges lose energy

    (Like loads )

    = Joules = N.m w

    Potential difference:

    Potential difference between two points is the energy gained or lost by a unit

    charge as it moves from on point to the other.

    coulombJoulevolt =

    =qwv

    w q= work done for transporting a total charge .

    Potential difference

    Voltage

    Potential

    -2-

  • Power :- Time rate for doing work .

    IVtq

    qw

    qq

    twP

    twP

    .=

    =

    =

    =

    Watt (w) = sJ = V.A

    Systems of unit ( S.I. ) :-

    Quantity Unit Symbol

    1) Charge (q) Coulombs C

    2) Current (I) Amper SCA =

    3) Force (F) Newton N

    4) Work , energy (w) Joule J=N.m

    5) Voltage (V) Volt cJV =

    6) Power (P) watt AVsJw .==

    7) Length ( l ) meter m

    8) Temperature (T) Kelvien K

    -3-

  • Electrical Circuits :-

    Circuit element :- is a two terminal electrical component .

    Electrical circuit :- Interconnected group of elements .

    Source = current and voltage in the same direction.

    load = current and voltage in the opposite direction .

    Resistive element, is always load and always dissipates power

    + VA

    I

    V

    - VB

    I

    + VA

    I

    V

    - VB

    I

    - VA

    I

    V

    + VB

    I

    + VA

    I

    V

    - VB

    I

    - VA

    I

    V

    + VB

    I

    (1) Load

    issipates power) P = V.I

    (2) Source

    (supplies power)

    (3) Source

    (supplies power , charging)

    (4) Load

    (dissipates power) (d

    +I

    V

    -

    -4-

  • Source of emf ( electro motive force ) s

    Source ( supplies power )

    Load ( consumes power )

    Ex. Charging of batteries.

    Resistance :-

    The unit of resistance ( R ) is ohm ( )

    The resistance of any material is depends on four factors:

    1. The length of conductor ( l ) lR 2. Cross section area of conductor ( A ) A

    R 1 3. The nature of material conductor.

    4. The temperature of conductor.

    AR l

    AR l= = constant is called resistivity or specific resistance unit of resistivity.

    -

    -

    V I

    I

    -

    l

    A

    -5-

  • mmmRA

    AR .==== l or

    .l

    2

    cm. 2

    2

    2==A dr

    r = radians of section.

    d = diameter of section.

    Con ctance ( G ):-du

    ll A

    AR

    === 11 (Siemens (S)) G

    Prefix :-

    pico 10 P -12

    nano 10-9 n

    micro 10-6

    milli 10-3 m

    centi 10-2 c

    deci 10-1 d

    Kilo 103 K

    Mega 106 M

    Giga 109 G

    Tera 1012 T

    Example: What is the resistance of 3 Km length of wire with cross section area

    6 mm2 and resistivity 1.8 cm .

    Solution:

    ( ) ( )( ) === 9106 6AR 10310108.1 326l

    -6-

  • Example: What is the resistance of 100 m length of copper wire with a

    diameter of (1 mm) and resistivity 0.0159 m .

    Solution:

    ( ) ( ) =

    02.210010 23

    6

    22 101

    2

    dA===

    0159.0100100159.0 6R l

    Effect of Temperature on a resistance :-

    slop = TR = constant =

    1

    1

    212 TTTT212

    TTRRRRRR

    =

    =

    Example: The resistance of material is 300 at 10Co, and 400 at 60Co. Find

    its resistance at 50 Co ?

    Solution:

    oCTTRR /2

    1060300400

    12

    12 ==

    = slop

    =+==

    ==

    =3803008080300

    40300

    10503002

    1

    1

    RRR

    RRTTRR

    -7-

  • Also from

    the above figure we can sea

    01

    12

    TR

    TTR

    = 02

    01

    1

    02

    2 00

    T

    TTR

    T =

    RT

    01

    02

    1

    2

    TTTT

    RR

    = , hence

    01

    022022 TTTTA ==

    l 101

    1TTTT

    A

    l

    ple: Aluminum conductor has resistance 0.25 at 10 Co . Find its

    resistance at 65 Co ?

    Solution:

    Exam

    ( )( )

    .31.024630125.0

    236102366525.02

    01

    0212

    01

    02

    1

    2

    =

    =

    =

    =

    =

    R

    TTTTRR

    TTTTR

    The following table illustrate the value of resistivity (

    R

    ) for different materials t 20 Co temperature.

    a

    Material at 20 Co , .m Silver

    Paper

    Mica

    1.63*10-8

    2.83*10-8

    20*10-8

    10*1010

    1011-1015

    *Copper

    Aluminum

    (1.72 - 1.77)*10-8

    Iron

    Material To , Co

    Silver

    Aluminum

    Iron

    -234

    .5

    -236

    -180

    Copper -234

    -8-

  • Example: Aluminum conductor with length of 75 cm and 1.5 mm2 cross

    section area. Find its resistance at 90 Co ?

    Solution:

    ( ) ( )

    === m15.14105.141105083.2 44

    ==

    AR

    105.110751083.2

    6

    28l

    . : 20Co

    ( )( )

    .236202369015.

    209015

    1

    2

    ++

    TTR

    r method

    18= m14=236 236 .142 =R

    01

    02

    = TTR

    Anothe

    ++=

    =2362023690

    209001

    02

    1

    2

    TTTT

    ( )

    =

    =

    ++

    ==

    0 8

    m

    AR

    15.14

    183.2

    105.1

    10752362023690

    90

    20

    6

    220

    90

    l

    -9-

  • Driving :

    ( )

    ( )

    +=

    +=

    +=

    +=

    0112 TT +=

    =

    =

    1201

    12

    01

    1212

    01

    010212

    01

    0212

    02

    01

    0212

    01

    02

    1

    2

    11

    1

    1

    11

    11

    TTTT

    RR

    TTTTRR

    TTTTTTRR

    TTTTR

    TTRR

    TTTTRR

    TTTT

    RR

    Let

    R

    011

    1TT = temperature coefficient of resistance at a temperature T1

    ( )[ ]12112 1 TTRR +=

    Where T0 for copper = -234.5

    ource, T0 take an absolute value, which means In some res 0T = 234.5, hence

    e can sea w

    & 10

    20

    1

    2

    TTTT

    RR

    ++=

    11

    1TT +=

    Example:

    a) Find the value of 1 at (T1 = 40 Co) for copper wire. b) Using the result of (a), find the resistance of a copper wire at 75 Co if its

    resistance is 30 at 40 Co ?

    -10-

  • -11-

    Solution:

    a) ( ) 00364.05.2741

    5.2344011

    011 ==== TT 1/K

    Or 00364.05.274

    1405.234

    11

    11 ==+=+= TT 1/K

    b) ( )[ ]12112 1 TTRR += ( )[ ] =+= 8.33407500364.0130

    .

    Example: If the resistance of a copper wire at freezing ( 0 Co ) is 30 , Find its

    sistance at -40 Co ?

    Solution:

    re

    ( )( )

    == 88.245.19430

    =

    =

    5.234

    5.23405.2344030

    01

    0212 TT

    TTR

    Or

    R

    + 20 TT +

    =101

    2

    TTRR

    == 88.245.23430

    5.194

    40 += 05.23430 5.234

  • Ohm's Law :- Ohm's law states that the voltages ( V ) across a resistor ( R ) is

    directly proportional to the current ( I ) flowing through the resistor .

    Slop = RV

    I 1=

    V = constant = R I

    IVR = ; V = I . R ;

    RVI =

    The resistance of short circuit element is approaching to zero. The resistance of open circuit is approaching to infinity.

    S.C.==

    =

    RG

    R10R

    Hence VI

    RG == 1 Siemens ( S ) or mhos ( ) .

    O.C.01 ==

    =G

    R

    -12-

  • Electrical Energy ( W ) :-

    tPWtWP .==Q KWh

    ( )( )

    tRV

    tRItIVtPW

    .

    .....

    2

    2

    =

    ==

    =

    Energy in KWh ( W ) = ( ) ( )1000

    ttimePPower

    Example : For the following circuit diagram , calculate the conductance and

    the power ?

    Solution :

    mARVI 6

    10530

    3 ===

    mR

    G 2.010511

    3 ===

    ( ) mWVIP 18030106. 3 === or ( ) ( ) mWRIP 180105106. 332 === or ( ) ( ) mWGVP 180102.030. 322 === or ( )( ) mWRVP 180102.0 30 3

    22

    ===

    -13-

  • Efficiency ( ) :-

    Wi/p = Wo/p + Wloss

    ttt+= WWW losspopi //

    Pi/p = Po/p + Ploss

    %100powerInput powerOutput Efficiency ( ) =

    %100=i

    o

    PP

    %100=i

    o

    WW

    nT = ......321

    Example: A 2 hp motor operates at an efficiency of 75 %, what is the power

    input in Watt, if the input current is (9.05) A, calculate also the input

    voltage?

    Solution:

    1 hours power (hp) = 746 Watt

    %100=i

    o

    PP

    -14-

  • WPP ii

    33.198975.0

    1492746275.0 ===

    VIPEIEP 22082.219

    05.933.1989. ====

    Example: What is the energy in KWh of using the following loads:-

    a) 1200 W toaster for 30 min.

    b) Six 50 W bulbs for 4 h.

    c) 400 W washing machines for 45 min.

    d) 4800 W electric clothes dryer for 20 min.

    Solution : ( ) ( )

    KWh

    W

    htWPW

    7.310003700

    100016003001200600

    100060204800

    60454004506

    60301200

    1000

    ==+++=

    +

    ++

    =

    =

    D.C. Sources:-

    The d.c. sources can be classified to:-

    1- Batteries . Voltage

    Amper - hours

    2- generators .

    3- Photo cells .

    4- Rectifiers .

    -15-

  • VE

    E

    V

    I

    V = E = constant voltage element

    V

    IIo

    IIo

    I = Io = constant current element

    .

    .

    Series Circuit :-

    V1 = I.R1

    V2 = I.R2

    V3 = I.R3

    E V1 V2 V3 = 0 E = V1 + V2 + V3 E = I.R1 + I.R2 + I.R3

    -16-

  • E = I.[R1 + R2 + R3] = I.RT

    The current in the series circuit is the same through each series element &

    RT = R1 + R2 + R3 + -------- + RN

    3

    3

    2

    2

    1

    1

    RV

    RV

    RV

    REIT

    ====

    Pt = P1 + P2 + P3 = E.I

    Voltage Source in Series:-

    -17-

  • Example: Find the current for the following circuit diagram?

    Solution:

    ET = 10 + 7 + 6 3 = 20 V

    RT = 2 + 3 = 5

    AREIIT

    TT 45

    20 ====

    Kirchoff's voltage law ( K.V.L. ):-

    The algebraic sum of all voltages around any closed path is zero.

    =

    =m

    mmV

    10

    Where m is the number of voltages in the path ( loop ) , and Vm is the mth

    voltage .

    E V = 0

    E = V

    RE

    RVI ==

    -18-

  • E V1 V2 = 0

    E = V1 + V2 ; RT = R1 + R2

    TT RVV

    REI 21 +==

    Example: Use K.V.L. to find the current in the following circuit diagram?

    E1 E2

    R1

    V1

    R2

    V2

    I

    Solution: From K.V.L. = 0V E1 V1 E2 V2 = 0

    E1 E2 = V1 + V2

    E1 E2 = IR1 + IR2

    E1 E2 = I ( R1 + R2 )

    21

    21

    RREEI +

    =

    -19-

  • Example: For the following circuit diagram, Find I using:-

    a) Ohm's law.

    b) K.V.L.

    7

    10

    40V

    10V

    20V10V

    17

    6

    Solution:

    a ) By applying ohm's law :-

    -20-

    AREI T 1

    4040

    17671010104020 ==+++

    +==

    b ) By applying K.V.L. :-

    10 + 6I + 7I - 40 + 10I 20 + 10 + 17I = 0

    10 40 20 + 10 + I ( 6 + 7 + 10 + 17 ) = 0

    -40 = -I ( 40 ) AI 14040 ==

  • Example :- For the following circuit diagram , find the potential difference

    between Node ( A & D ) , and Node ( A & F ) ?

    5V

    12V8V

    6V

    C F

    DA

    B E

    Solution : To find the potential difference between Node A & D , we will apply

    K.V.L. on the closed loop BADEB

    5V

    12V8V

    6V

    C F

    DA

    B E

    +6 + V 12 8 = 0 V2

    V1

    VV = 20 6 = 14 volt

    or

    +6 V1 12 8 = 0

    14 V1 = 0

    V1 = 14 volt * D

    A 14 .

    Take the loop FEDAF to find the potential difference between Node C & F .

    5 + 12 V V2 = 0 5 + 12 14 V2 = 0 7V2 = 0 V2 = 7 volt . Or Take the loop FEBAF 5 8 +6 V2 = 0 V2 = 7 volt .

    -21-

  • Example :- For the following circuit diagram , find the current ?

    Solution :

    315V

    8V

    22

    2

    2

    222

    2

    IV

    A B C

    F E D Take the loop FABCDEF

    +8 + V 15 = 0 +V 7 = 0 V = +7 volt

    V = IR AI 5.327 ==

    Definitions :-

    Node :- Meeting point of 3 or more branches .

    Branch :- Series of elements carrying the same current .

    Loop :- Is any closed path in a circuit .

    -22-

  • Hence for the loop circuit, we can find :-

    A4 nodes and 6 branches

    E1 E2

    V3 V4

    and we can find : V1 , V2 , V3 and V4

    as follows :- BD

    C

    E3

    V1

    V2

    Take the loop BACB ; to find V1

    E2 V1 E3 = 0 V1 = E2 E3 Or, if we take BCAB ;

    E3 V1 E2 = 0 V1 = E2 E3 Take the loop ADBA ; to find V2

    E1 + V2 + E2 = 0 V2 = E1 E2

    Take the loop ABCDA ; to find V3

    E2 + E3 + V3 + E1 = 0

    V3 = E2 E3 E1

    V4 = V3

    Or;

    E2 + E3 V4 + E1 = 0

    V4 = E1 + E3 E2

    Example :- For the following circuit diagram , find ; RT , I , V1 , V2 , P4 , P6

    , PE , verify by K.V.L. ?

    R1 R2

    V1 V2

    I

    E=20V

    4 6

    Solution :-

    RT = R1 + R2 = 4 + 6 = 10

    AREIT

    21020 ===

    -23-

  • VIRV 84211 === VIRV 126222 ===

    ( ) WRIP 1642 2124 === ; or ( ) wRVP 16

    48 2

    1

    21

    4 ===

    ( ) WRIP 2462 2226 === ; or ( ) wRVP 24

    612 2

    2

    22

    6 ===

    WIEPE 40202 === ; or WPPPE 40241664 =+=+= To verify results by using K.V.L. ; then

    01

    ==

    N

    iiV

    E V1 V2 = 0

    E = V1 + V2

    20 = 8 + 12

    20 = 20 checks

    Internal Resistance :-

    Every practical voltage or current source has an internal resistance that

    adversely affects the operation of the source.

    In a practical voltage source the internal resistance represent as a resistor in series

    with an ideal voltage source.

    In a practical current source the internal resistance represent as a resistor in

    parallel with an ideal current source, as shown in the following figures.

    RL

    E

    I

    VVo

    Practical voltage source

    RoRoI

    Practical current source

    RL

    -24-

  • Where

    Ro = Internal resistance

    RL = load resistance

    According to K.V.L.

    E Vo V = 0

    E IRo V = 0

    V = E IRo Note that an ideal sources have Ro = 0

    We can representing a load as a group of parallel resistances.

    Hence as the load will increase the current will be increase ( because the

    resistance will decrease ) and the voltage will decrease .

    This is because the drop voltage due to the internal resistance , as shown in the

    following figure :-

    -25-

  • As seen from the above figure , if Ro2 > Ro1 , then V2 < V1 and the drop voltage

    will be ( E V2 ) , which is greater than ( E V1 ) .

    Example :- For the following circuit diagram , calculate I and VL for the

    following cases :-

    a) Ro = 0

    b) Ro = 8

    c) Ro = 16

    Solution :-

    a.) By apply K.V.L.

    E Vo V = 0

    120 IRo IRL = 0 120 0 22I = 0 120 = 22I I = 5.46 A VL = I RL = 5.46 * 22 = 120 V

    b.) E Vo V = 0

    120 8I 22I = 0 120 30I = 0 120 = 30I I = 4 A VL = I RL = 4 * 22 = 88 V

    c.) E Vo V = 0

    120 16I 22I = 0 120 38I = 0 120 = 38I I = 3.16 A VL = I RL = 3.16 * 22 = 69.5 V

    Then we can conclude that as Ro increase the total current and load voltage will

    decrease.

    -26-

  • Example :- A circuit have load one with 20 and 4A , and load two with 10

    & 6A . Find the current for load three which have 30 ?

    Solution :-

    1) 20 & 4A

    2) 10 & 6A

    3) 30 & I = ?

    From K.V.L. , then

    E Vo VL = 0

    VL = E IRo

    IRL = E IRo

    4 * 20 = E 4Ro

    80 = E 4Ro

    ----------------- ( 1 )

    Also

    6 * 10 = E 6 Ro ----------------- ( 2 ) 60 = E 6Ro

    From eq. (1) & (2) , we have

    20 = ( 6 4 ) Ro Ro = 10 ; sub. this result it in eq. (1) , then 80 = E 4 * 10 E = 120 V Now , we Apply K.V.L. for load 3 ;

    V3 = E IRo 30I = 120 10I 40I = 120 I = 3 A for load three. See from this example that the current will increase as the load will decrease

    with constant E & Ro .

    -27-

  • Example :- A circuit have Voc = 25 v and Isc = 50 A , find its current and RL

    when VL = 15 V ?

    Solution :-

    E = Voc = 25 V

    === 5.05025

    sco I

    ER

    From K.V.L.

    E Vo VL = 0

    E 0.5I 15 = 0

    25 0.5I 15 = 0

    0.5I = 25 15

    -28-

    AI 205.0

    10 ==

    === 75.02015

    IVR LL

  • Voltage divider Rule :-

    RT = R1 + R2

    TREI =

    TT RRER

    RERIV 1111

    ... =

    ==

    TRRER

    RERIV 22

    222

    ... =

    ==

    Voltage divider rule

    Vn = Voltage across Rn

    E = The ( emf ) voltage across the series elements .

    RT = The total resistance of the series circuits .

    Example :- Using voltage divider rule , determine the voltage V1 , V2 , V3 and

    V4 for the series circuit in figure below , given that ; R1 = 2K , R2 = 5K ,

    R3 = 8K , E = 45 V ?

    Solution :-

    R1 R2

    V1 V2

    I

    E

    R3

    V3

    V4

    VRERVT

    1510*15

    45*10*53

    32

    2 ===

    T

    nn R

    ERV =

    VRERVT

    610*15

    45*10*23

    31

    1 ===

    -29-

  • VRERVT

    2410*15

    45*10*83

    33

    3 ===

    ( ) VR

    ERRVT

    2110*15

    45*10*73

    321

    4 ==+= or V4 = V1 + V2 = 21V

    To check: E V1 V2 V3 = 0

    E = V1 + V2 + V3 45 = 6 + 15 + 24 45 = 45

    Active Potential :-

    a

    b

    Va = 14 V

    Vb = 8 V

    Vab = 6 V

    Vab is the voltage difference between the

    point a and point b

    Vab = Va Vb = 14 8 = 6V

    Vba = Vb Va = 8 14 = - 6V

    Vab = - Vba

    a

    -7 V

    a

    10 V

    a

    Va = -7 V

    Va = 10 V

    Va = 0 V

    -30-

  • R1

    R2

    R1

    R2

    20 V

    E = 20 V

    R1

    R2

    -12 V

    E = -12 V

    R1

    R2

    -31-

  • Example :- Find Va , Vb , Vc , Vab , Vac and Vbc for the following diagram .

    Solution :-

    RT = R1 + R2 + R3

    = 2 + 5 + 3 = 10

    AREIT

    11010 ===

    E V2 Va = 0

    Va = E V2 = 10 (2 * 1) = 8 V

    Vb = V5 = (1 * 5) = 5 V = Vbc ; Vc = 0 V

    or E V2 V3 Vb = 0 Vb = E V2 V3 = 10 2 3 = 5 V Vab = Va Vb = 8 5 = 3 V

    Vac = Va Vc = 8 0 = 8 V

    Vbc = Vb Vc = 5 0 = 5 V

    -32-

  • Equivalence of actual sources :-

    Voltage

    Source

    Current

    Source

    Open

    Circuit

    Voc = E

    I = 0 oooc GIV 1=

    Short

    circuit osc R

    EI =

    V = 0

    Isc = Io

    Kirchoff's Current Law ( K.C.L. ) :-

    The algebraic sum of ingoing currents is equal to the out going currents

    at any point .

    = outin IIOr , At any point , the algebraic sum of entering and leaving current is zero .

    = 0I

    I1

    I5

    I2I3

    I4

    I1 + I2 + I4 = I3 + I5

    Or I1 + I2 + I4 - I3 - I5 = 0

    -33-

  • At a

    I1 = I2 + I3 13 + 5 I = 0

    Or I1 - I2 - I3 = 0 18 I = 0

    At b I = 18 A -I1 + I2 + I3 = 0

    Example :- Find the current in each section in the cct. Shown ?

    a

    b

    cd

    e 3A

    4A8A

    2A

    3A

    Iab

    Icd

    Ibc

    Ide

    1A

    Solution :-

    At node a

    3 1 Iab = 0

    2 Iab = 0 Iab = 2 A

    -34-

  • At node b

    Iab + 3 Ibc = 0

    2 + 3 Ibc = 0 Ibc = 5 A At node c

    Ibc + 4 Icd = 0

    5 + 4 Icd = 0 Icb = 9 A At node d

    Icd 8 Ide = 0

    9 8 Ide = 0 Ide = 1 A At node e

    2 3 + Ide = 0

    2 3 + 1 = 0

    0 = 0 check .

    Example :- Find the magnitude and direction of the currents I3 , I4 , I6 , I7 in the

    following cct. Diagram?

    a

    b

    c

    dI1 = 10A

    I 2= 12

    AI5 = 8A

    I7

    I6I3

    I4

    Solution :-

    = leaveenter II I1 = I7 = 10 A

    -35-

  • At node a ; suppose I3 is entering

    I1 + I3 I2 = 0

    10 + I3 12 = 0 I3 = 2 A At node b;

    I2 enter , I5 leave , I4 must be leaving I2 = I5 + I4

    12 = 8 + I4 I4 = 12 8 = 4 A At node c;

    I4 enter , I3 leave , I6 leave I4 = I3 + I6

    4 = 2 + I6 I6 = 2 A At node d;

    I5 and I6 enter , I7 leave

    I7 = I5 + I6

    10 = 8 + 2

    10 = 10 Ok.

    Resisters in Parallel :-

    R1 R2

    I2I1

    V1 V2

    I

    V

    A

    B

    From K.V.L. V = V1 = V2

    From K.C.L. I = I1 + I2

    2

    2

    1

    1

    RV

    RVI +=From .L.

    = V1G1 + V2G2 = V1 ( G1 + G2 )

    or I = V ( G1 + G2 )

    I = VGT

    -36-

  • Where GT = G1 + G2

    Hence 21

    21

    21 .111

    RRRR

    RRRT

    +=+=

    or 21

    21.RRRRRT +=

    In the same minner , if we have three resistors in parallel , then:

    321

    1111RRRR T

    ++=

    321

    213132

    .....1

    RRRRRRRRR

    RT

    ++=

    213132

    321

    .....

    RRRRRRRRRRT ++=

    And , if we have N of parallel

    resistance , then

    NT RRRRR11111

    321

    +++=

    Also

    PT = P1 + P2 + P3

    1

    21

    12

    1111 RVRIIVP ===

    Source power T

    TTTTs R

    ERIEIP2

    2 ===

    -37-

  • Example :- For the following cct. Find RT , PT , IT , Ib?

    816 V

    IT

    8 8 8

    Solution :-

    === 248

    NRRT

    AREIT

    T 8216 ===

    AREIbranch 28

    16

    1

    ===

    ( ) ( ) WRIP TTT 1282.8 22 === or PT = E.IT = 16 * 8 = 128W

    or PT = P1 + P2 + P3 + P4

    ( ) ( ) ( ) ( )

    W12832323232

    8*28*28*28*2 2222

    =+++=

    +++=

    -38-

  • Example :- For the parallel network in fig. below , find :-

    a) R3 , b) E , c) IT , I2 , d) P2 ; given that RT = 4 ?

    Solution :-

    a)

    ===

    =

    ++=

    ++=

    ++=

    101.0

    111.0

    105.01.025.0

    105.01.025.0

    1201

    101

    41

    1111

    33

    3

    3

    3

    321

    RR

    R

    R

    R

    RRRRT

    b) E = V1 = I1R1 = 4 * 10 = 40 V

    c)

    ARE

    RVI

    AREIT

    T

    22040

    104

    40

    22

    22 ====

    ===

    d) ( ) ( ) WRIP 8020.2 22222 ===

    or 2

    22

    2 RVP = , or P2 = I2V2

    -39-

  • Current division Rule :-

    21

    21.RRRRIV +=

    1

    21

    21.

    RRRRRI

    RVIT

    +==

    21

    21 RR

    RII +=

    In the same miner

    21

    12 RR

    RII +=

    Also 1

    2

    21

    1

    21

    2

    2

    1

    RR

    RRRI

    RRRI

    II =

    +

    +=

    2

    1

    1

    2

    2

    1

    GG

    RR

    II ==

    Example :- For the following circut. , find V , I1 and I2?

    Solution :-

    ==+=+= 0999.01.10010

    1.01001.0*100.

    21

    21

    RRRRRT

    V = I . RT = 5 * 0.0999 = 0.4995 V

    -40-

  • AVI 004995.01001

    ==

    AVI 995.41.02==

    To check I = I1 + I2

    5 = 0.004995 + 4.995

    5 = 5 Ok.

    Example :- Determine the resistance R1 in the figure below?

    Solution :-

    I = I1 + I2

    or I2 = I I1 = 27 21 = 6 mA

    V2 = I2R2 = 6 * 10-3 * 7 = 42 mV

    V1 = V2 = 42 mV

    ===

    210*2110*42

    3

    3

    1

    11 I

    VR

    or

    =

    +=+=

    2

    77*10*2710*21

    1

    1

    33

    21

    21

    R

    RRRRII

    -41-

  • Voltage Regulation :-

    Voltage Regulation %100% =FL

    FLNLR V

    VVV

    Where

    VNL = No load voltage

    VFL = Full load voltage

    Also we can write

    %100% .int =L

    R RRV

    Where

    Rint. = Internal resistor .

    RL = load resistor .

    Example :- Find the voltage VL and power lost to the internal resistance , if the

    applied load is 13 , also find the voltage regulation ?

    Rint. = 2

    RL = 13VLE = 30V

    Solution :-

    ARR

    EIL

    L 213230

    int

    =+=+=

    VRIEV LL 262*230.int === ( ) ( ) WRIP Lloss 82.2 2.int2 === %385.15%100

    262630%100% ===

    FL

    FLNLR V

    VVV

    or %385.15%100132%100% .int ===

    LR R

    RV

    -42-

  • Example :- Find the current I , for the network shown:

    R1 = 6

    I1

    I = 42 mA

    R2 =24 R3 =24

    Solution :- All resistance in parallel , so if we define that R = R2 // R3 then :-

    =+=+= 12242424*24

    32

    32

    RRRRR

    Hence

    ( ) mARR

    RII 28612

    1210*42 31

    1 =+=+=

    Example :- Calculate I & V for the network shown

    Solution :- We have a short circuit on R2 resistance , hence no current through

    R2 , hence the above cct. Can redrawn as fellows:

    -43-

  • mARE

    REIT

    6.310*5

    183

    1

    ====

    VERIV 18. 1 ===

    Example :- For the following cct. Network , find RT , IA , IB , IC , VA , VB , I1 ,

    I2 ?

    R1 = 9

    R2 = 6

    R5 = 3R4 = 6

    R3 = 4

    R6 = 316.8 V

    IA

    I1

    I2 IB IC

    Solution :-

    RC = 3

    IC

    IB

    IA

    RB = 6

    RA = 3.6=+=+= 6.3696*9

    21

    21

    RRRRRA

    =++= 6363*94RB = R3 + R4 // R5

    RC = 3

    RT = RA + RB // RC

    =++= 6.5363*66.3

    ARE

    TA 36.5

    8.16 === I

    -44-

  • Apply C.D.R.

    ARR

    RIICB

    CAB 163

    3*3 =+=+=

    By K.C.L.

    IC = IA IB = 3 1 = 2 A

    VA = IARA = 3 * 3.6 = 10.8 V

    VB = IBRB = 1 * 6 = 6 V = VC

    ARRRII A 2.1

    963*6

    21

    21 =+=+=

    I2 = IA I1 = 3 1.2 = 1.8 A

    To check

    E VA VB = 0

    16.8 10.8 6 = 0

    0 = 0 Ok.

    -45-

  • Example :- Find the resistor required to connect in parallel with the ammeter to

    flow 1.2 A , if you know that the fsd ( full scale deflection ) of ammeter is 120

    mA , and the resistance of ammeter is 2.7 ?

    Solution :-

    From K.C.L.

    Ish = 1.2 0.12 = 1.08 A

    +=+=

    sh

    shA

    Ash

    R

    RRRII

    7.27.22.108.1

    ( )

    =

    =

    +=+=

    +=

    3.008.1324.0

    08.1916.224.37.208.124.3

    7.224.308.1

    sh

    sh

    sh

    sh

    sh

    RR

    RR

    R

    -46-

  • Example :- for the following cct. Network , Given that (V= 24 v), Find E ?

    8

    E

    24

    8

    4

    4

    1612

    6

    6

    Solution :-

    AI 22.1

    241 == V

    The same voltage ( V = 24 V ) on the

    resistor Ra = 4 +16 + 4 = 24

    Hence

    AI 12424

    2 ==

    AIII 3213 =+=Also from K.C.L. I4 = I3 = 3A

    Take the closed loop ABCDA , from

    K.V.L.

    V1 6I4 V 6I3 = 0

    8

    E

    24

    8

    4

    4

    1612

    6

    6

    V

    V1

    V2

    I5B

    V3C

    I6

    I4

    D

    I2AI3I1

    V1 = 6 * 3 + 24 + 6 * 3

    VV 601 =

    AI 5.22460

    5 ==

    AIII 5.535.2456 =+=+=Take the closed loop CBC

    V1 V2 + E V3 = 0

    E = V1 + V2 + V3

    E = 60 + 25.5 * 8 + 5.5 * 8

    = 60 + 44 + 44

    E = 148 V

    -47-

  • Current Source :-

    Example :- Find the voltage ( Vs ) for the circuit below:

    10 A

    Solution :-

    I RL (2-5)Vs = IRL = 10 * 2 = 20 V if RL = 2 VsVs = IRL = 10 * 5 = 50 V if RL = 5

    Example :- Calculate V1 , V2 , Vs for the following cct.:

    Solution :-

    R1

    R2

    VsI = 5A

    V1

    V2

    2

    3

    V1 = IR1 = 5 * 2 = 20 V

    V2 = IR2 = 5 * 3 = 15 V

    Vs = V1 + V2 = 10 + 15 = 25 V

    Source Conversions :-

    A voltage source with voltage E and series resistor Rs can be replaced by

    a current source with a current I and parallel resistor Rs as shown :-

    sREI =

    Current source to voltage source

    Voltage source to current source

    -48-

  • Example :- Convert the voltage source in the cct. Below to a current source ,

    then calculate the current through the load for each source:

    ILSolution :-

    Rs = 2

    RL = 4A

    RREI

    LsL 142

    6 =+=+= E = 6V

    For the current source cct. A

    RRRII

    Ls

    sL 142

    23 =

    +=+=

    RL = 4

    IL3A

    Rs = 2

    A

    REIs

    326 ==

    =

    Example :- Convert the current source in the cct. Shown below to a voltage

    source and determine IL for each cct.:

    Solution :-

    IL

    .

    For the current cct. ( )

    mAIRR

    RII

    L

    Ls

    sL

    310*610*3

    10*310*9 333

    3

    =

    +=+=

    For the voltage source cct.

    ( )mAI

    RRE

    REI

    L

    LsTL

    310*63

    273

    =+=+==

    -49-

  • Current source in parallel :-

    Is = 10 6 = 4 A & Rs = 3 // 6 = 2

    Example :-

    Is = 7 3 + 4 = 8 A

    Example :- Find the load current in the following cct.:

    Solution :-

    -50-

  • AREI 4

    832

    11 ===

    Is = I1 + I2 = 4 + 6 = 10 A

    Rs = R1 // R2 ARRRII

    Ls

    ssL 3146

    6*10624824*8 =+=+==+=

    -51-

  • Matrices :-

    Second order determinate

    Col. 1 Col. 2

    -

    122122

    1 babaa

    baD == 1

    b

    Col. 1 Col. 2 Col. 3

    a1x b1y = c1

    a2x b2y = c2

    =

    2

    1

    22

    11

    cc

    yx

    baba

    2121

    2121

    22

    11

    22

    11

    1

    abbacbbc

    bababcbc

    DDx

    =

    ==

    2121

    2121

    22

    11

    22

    11

    2

    abbaacca

    babacaca

    DDy

    =

    ==

    Example :- Find the value of D

    =2614

    D

    Solution :-

    ( ) 14686*12*42614 =+==

    =D

    -51-

  • Example :- Solving the equations below by determinates

    4I1 6 I2 = 8

    2I1 + 4 I2 = 20

    Solution :-

    =

    208

    4264

    2

    1

    II

    ( )( ) ADDI 428.5

    121612032

    2*64*420*64*8

    426442068

    11 =+

    +==

    ==

    ADDI 28.2

    282*820*4

    2820284

    11 ==

    ==

    Third order determinant :-

    333

    222

    111

    cbacbacba

    3

    2

    1

    aaa

    3

    2

    1

    bbb

    D =

    [ ] [ ]321321321321321321 cabbcaabcbacacacbaD ++= + +

    Example :- Find the value of D

    =

    240012321

    D

    -52-

  • Solution :-

    412

    02

    1

    240012321

    =D

    ( )[ ] [ ][ ] [ ] 148228002402

    2*2*21*0*43*1*04*2*30*0*22*1*1

    =+=+++=

    D

    D ++= + +

    Example :- Find V1 , V2 , V3 from the following equations :-

    2V1 + 4V2 +2V3 = 8

    5V1 2V2 10V3 = 18

    V1 + 8V2 20V3 = -8

    Solution :-

    =

    8188

    20811025242

    3

    2

    1

    VVV

    82

    4

    152

    20811025242

    82

    4

    8188

    208810218248

    11

    ==DDV

    ( ) ( ) ( ) ( )[ ] ([ ) ( ) ( ) ( ) ]

    ( ) ( ) ( )[ ] ( ) ( ) ( )[ ]4*5*202*10*82*2*18*5*21*10*420*2*24*18*208*10*82*2*88*18*28*10*420*2*8

    1 +++++=V + + +

    VV 35.46842976

    1 ==

    -53-

  • 6848

    188

    152

    208110185282

    22

    ==DDV

    68482

    4

    152

    8811825842

    33

    ==DDV

    Star Delta ( ) and Delta Star ( ) transformation :-

    1. ) Delta Star ( ) transformation :- If the value of RAB , RCA , RBC are

    known, and we need to get the values

    of RA , RB , RC ; then :-

    If RAb = RBC = RCA = R , in this case ==== RRRRR CBA 3

    or 3

    = RR

    BCCAAB

    CAABA RRR

    RRR ++=

    BCCAAB

    BCABB RRR

    RRR ++=

    BCCAAB

    BCCAC RRR

    RRR ++=

    -54-

  • BA

    C30

    3030

    B

    A

    C

    10

    10 10

    2. ) Star Delta ( ) transformation :-

    If the value of RA , RB , RC are known ,

    and we need to get the values of RAB ,

    RCA , RBC ; as follows :-

    If === RRRR CBA , in this case RAB = RCA = RBC = R = 3 RY

    B

    A

    C

    4

    4 4

    B

    A

    C12

    1212

    C

    CACBBAAB R

    RRRRRRR ++=

    A

    CACBBABC R

    RRRRRRR ++=

    B

    CACBBACA R

    RRRRRRR ++=

    -55-

  • Examples of star and delta connections and transformation :-

    Delta connection Star connection

    Example :- Find the current flow in the 25 V source for the following circuit :-

    Solution :-

    =++= 5.21510515*5

    aR

    =++= 51510515*10

    bR

    =++= 67.11510510*5

    cR

    25 V

    Rc Rb Ra

    5 8 10

    { ( ) ( ) } ( ){ ( ) ( ) } ( ){ }

    AREI

    R

    R

    RR

    RRRR

    T

    T

    T

    T

    T

    cbaT

    92.104.13

    2504.1367.637.6

    67.6135.1213*5.12

    67.613//5.12567.185//105.2

    58//10

    ====+=

    +

    +=

    +=++++=++++=

    -56-

  • Example :- For the following network , find I ?

    Solution :- The resistances ( 6 , 3 , 3 ) are delta , can convert to star connection

    as follows:

    =++= 5.13363*6

    aR , =++= 5.13363*6

    bR , =++= 75.03363*3

    cR

    ( ) ( )[ ]( ) ( )[ ]

    75.05.35.55.3*5.5

    5.3//5.52//4

    +

    +=

    +=+++=

    c

    cbaT

    RRRRR

    AREIT

    077.2889.26 ===

    -57-

  • Example :- Find I for the following cct. network :-

    12 V

    5

    2015

    12.5 10

    30

    I

    Solution :- The resistors ( 5 , 10 , 20 ) are star convert to delta

    I

    15

    12.5

    30

    b

    a

    c

    Rbc

    Rab

    Rac

    12 V

    =++=

    =++=

    =++=

    3510

    20*55*1020*10

    5.1720

    20*55*1020*10

    705

    20*55*1020*10

    bc

    ac

    ab

    R

    R

    R

    It is clear that ( 12.5 // Rac ) and (15 // Rbc ) and (30 // Rab ) , hence the

    circuit can be reduce to the following network :-

    -58-

  • 12 VR3

    I

    R1

    R2

    ( )

    -59-

    =+== 3.75.175.125.17*5.12//5.121 acRR

    ( ) =+== 5.10351535*15//152 bcRR

    ( ) =+== 21703070*30//303 abRR

    ( )( )

    =+==+=

    +=

    634.9218.1721*8.1721//8.17

    21//5.103.7

    // 321 RRRRT

    AREIT

    246.1634.912 ===

  • Network Solution :-

    To solve a circuit is to find the current and voltage in all branches.

    1) Loop ( Mesh ) current method :-

    Example( 1 ):- Find the current through the 10 resistor of the network

    shown:

    Solution :-

    - * Loop + * Loop m

    ) Loop = ( .

    The loop equations are :-

    Loop 1 :-

    - ( 8+3 )I1 + 3I2 +8I3 + 15 = 0

    Loop 2 :-

    - ( 3+5+2 )I2 + 3I1 +5I3 = 0

    Loop 3 :-

    - ( 10+8+5 )I3 + 8I1 +5I2 = 0

    Rearrange the equations , then :-

    -11I1 + 3I2 +8I3 = -15

    3I1 - 10I2 +5I3 = 0

    8I1 + 5I2 - 23I3 = 0

    -60-

  • AII

    ADDI

    22.1

    22.1

    235851038311058010315311

    103

    33

    ==

    =

    ==

    Example( 2 ):- Solve following circuit diagram;

    Solution :-

    -I1 ( 5+7 ) + 7I2 + 20 5 = 0

    -I2 ( 7+2+6 ) + 7I1 + 6I3 + 5 + 5 + 5 = 0

    -I3 ( 6+8 ) + 6I2 5 30 = 0

    Rearrange;

    -12I1 + 7I2 +0 = -15

    7I1 - 15I2 +6I3 = -15

    0 + 6I2 - 14I3 = 35

    ------------------- ( 1 )

    ------------------- ( 2 )

    ------------------- ( 3 )

    -61-

  • ADDI 862.1

    14022610

    14606157071214635615150715

    11 ==

    ==

    ADDI 049.1

    14021470

    140214350615701512

    22 ==

    ==

    ADDI 05.2

    14022875

    140235601515715712

    33 ==

    ==

    Example( 3 ):- Find the current in the 10V source , for the following network;

    10V

    4 3

    6 I2I1

    5A

    Solution :-

    I2 = -5 A

    Hence , we need only one equation to solve this circuit

    -I1 ( 4+6 ) + 6 * ( -5 ) + 10 = 0

    -10I1 20 = 0 -10I1 = 20

    AI 210

    201 ==

    -62-

  • Example( 4 ):- Solve the following circuit diagram, also find the voltage across

    15 resistance?

    Solution:-

    I4 = 2 A

    -I1 ( 4+6+5 ) + 4I2 + 12 18 = 0

    -I2 ( 8+3+4+7 ) + 4I1 + 8I3 - 12 = 0

    -I3 ( 15+2+8+9 ) + 8I2 + 15 * 2 = 0

    Rearrange:-

    -15I1 + 4I2 +0 = 6

    4I1 - 22I2 +8I3 = 12

    0 + 8I2 - 34I3 = -30

    6O

    5O

    8O

    3O

    18V

    4O12V

    7O

    15O

    2O 9O

    2A

    I1

    I2

    I3

    I4

    ------------------- ( 1 )

    ------------------- ( 2 )

    ------------------- ( 3 )

    -63-

  • DDI 11 =

    DDI 22 =

    DDI 33 =

    V15 = I15 * R15

    = ( I3 I4 ) * 15

    = ( I3 2 ) * 15

    Example( 5 ):- Solve the following circuit diagram .

    3 15

    40

    5

    6020A

    10A

    5A

    Solution:- The above diagram can be reduced to the following diagram;

    -64-

  • 3 15

    40

    5

    60

    50V

    75V60V

    I1

    I2

    -I1 ( 5+15+60 ) + 60I2 - 50 75 = 0

    -I2 ( 3+60+40 ) + 60I1 + 60 = 0

    Rearrange:-

    -80I1 + 60I2 = 125

    60I1 - 103I2 = -60

    ------------------- ( 1 )

    ------------------- ( 2 )

    DDI 11 =

    DDI 22 =

    Example( 6 ):- Solve the following circuit diagram:

    10

    42

    6

    20V

    I1 I2

    6AVo

    -65-

  • Solution:-

    -( 6+2 )I1 + 2I2 +20 - Vo = 0

    -( 10+2+4 )I2 + 2I1 + Vo = 0

    I2 - I1 = 0

    Add eq. 1 & eq. 2

    -8I1 + 2I2 +20 -16 I2 + 2I1 = 0

    -6I1 - 14I2 = -20

    From eq. 3

    I1 I2 = -6

    ADDI 2.3

    1468420

    11146

    161420

    11 =+

    =

    ==

    ADDI 8.2

    202036

    2061206

    22 =+=

    ==

    ------------------- ( 1 )

    ------------------- ( 2 )

    ------------------- ( 3 )

    ------------------- ( 1 )

    ------------------- ( 2 )

    -66-

  • -: dohtem tnerruc pool gnisu , tiucric gniwollof eht evloS : ) 7 ( elpmaxE

    6

    8

    V21

    7

    9

    bI

    cI

    oV A3

    aI 5

    :noituloS

    -: a pooL 0 = oV - cI8 + aI) 8+5 (- ) 1 ( -------------------

    -: b pooL 0 = 21 oV + cI7 + bI) 7+6 (- ) 2 ( -------------------

    -: c pooL 0 = bI7 + aI8 + cI) 9+7+8 (- ) 3 ( -------------------

    3 = aI bI ) 4 ( -------------------

    oV -: . oV

    ) oV

    oV ( oV

    .

    -76-

  • Loop a+b :

    -13Ia - 13Ib + 15Ic 12 = 0

    Loop c :-

    -24Ic + 8Ia + 7Ib = 0

    Ib Ia = 3

    Rearrange Eq.s :-

    -13Ia - 13Ib + 15Ic = 12

    8Ia + 7Ib -24Ic = 0

    Ia Ib = 3

    ADDIa 862.11402

    2610

    3112478

    1513133102470

    151312

    1 ==

    ==

    DDIb 2=

    DDIc 3=

    ------------------- (1 )

    ) ------------------- ( 2------------------- ( ) 3

    ------------------- (1 ) ------------------- ( ) 2

    ) ------------------- ( 3

    -68-

  • Example, (Sheet 4 Q. 25): Solve the following circuit diagram using loop

    current:

    30

    15

    13V

    7

    9V

    3

    12

    1.2A

    0.8A

    20

    6V

    Solution:-

    15 // 30 = =+ 10301530*15 ; 12 + 3 = 15

    13V

    7

    15

    1.2A

    10

    6V

    9V

    16V

    20

    Ib

    Ic

    Ia

    Loop a:-

    -( 15+7 )Ia + 6 + 9 + 7Ib - 13 = 0

    ------------------- ( 1 )

    -69-

  • Loop b:-

    -( 7+10 )Ib - Vo + 13 + 7Ia = 0

    Loop c:-

    -20Ic 16 6 + Vo = 0

    Ic Ib = 1.2

    Loop b+c:

    -17Ib - 20Ic + 7Ia 9 = 0

    Rearrange Eq.s :-

    Loop a: -22Ia + 7Ib = -2

    Loop b+c: 7Ia - 17Ib -20Ic = 9

    Ib Ic = -1.2

    ------------------- ( 2 )

    ------------------- ( 3 )

    ------------------- ( 4 )

    ------------------- ( 1 ) ------------------- ( 2 )

    ------------------- ( 3 )

    Example, (Sheet 4 Q. 7): Solve the following circuit diagram:

    -70-

  • Solution:

    120

    80

    5

    2 10V

    1

    10 2

    8V20V

    4V10V

    Ib

    Ia

    Loop a :-

    -( 10+120+2+80 )Ia + 80Ib - 8 + 20 = 0

    Loop b :-

    -( 2+5+80+1 )Ib + 80Ia - 4 - 10 + 10 = 0

    Rearrange Eq.s :-

    -212Ia + 80Ib = -12

    80Ia - 88Ib = 4

    ------------------- ( 1 )

    ------------------- ( 2 )

    ------------------- ( 1 )

    ------------------- ( 2 )

    -71-

  • Nodal voltage:-

    Example 1 :- Solve the following circuit using the nodal voltage method:

    R4

    R1

    E2

    R2R3

    R5 R6

    E1

    VB - VCVA - VB

    VC

    VA

    VB

    A C

    D

    B

    VC - VA

    I1

    I4

    I6I5

    I3

    I2

    Solution :-

    Choose reference point N = 4

    Let D be a reference point IN = N-1 = 3

    Kcl at B:

    I5 I2 I6 = 0

    (VA VB ) G5 ( VB E2 ) G2 ( VB VC ) G6 = 0

    Kcl at A:

    I3 I5 I1 = 0

    -VA G3 ( VA VB ) G5 [( VA VC )- E1] G1 = 0

    -72-

  • Kcl at C:

    I6 + I1 I4 = 0

    ( VB VC ) G6 + [( VA VC )- E1] G1 VC G4 = 0

    Rearrange:

    A : ( VB VA ) G5 -VA G3 + ( VC VA ) G1 - E1G1 = 0

    B : (VA VB ) G5 + ( VC VB ) G6 - VB G2 + E2 G2 = 0

    C : ( VB VC ) G6 VC G4 + ( VA VC ) G1 - E1G1 = 0

    Hence, we can arrange the above equations in the following form:-

    A : - VA ( G1 + G3 + G5 ) + VBG5 + VCG1 + E1G1 = 0

    B : - VB ( G2 + G5 + G6 ) + VAG5 + VCG6 + E2G2 = 0

    C : - VC ( G1 + G4 + G6 ) + VAG1 + VBG6 - E1G1 = 0

    Then, we can find VA , VB , VC by the determinate method .

    Example 2 :- Solve the following circuit diagram using nodal voltage .

    Solution:

    First we simplify the circuit and make a less nodal point.

    ----------- ( 1 )

    ----------- ( 2 )

    ----------- ( 3 )

    -73-

  • N = 4 ; IN = 4 1 = 3

    Let D be a reference point

    A : 01935

    830

    191

    81

    71

    71

    81

    191 =++

    ++

    +

    ++ CBA VVV

    B : 033351

    71

    331

    71 =+

    +

    + AB VV

    C : 01935

    830

    81

    191

    191

    81

    61 =

    ++

    ++ AC VV

    Rearrange

    -0.321 VA + 0.143 VB + 0.178 VC = -5.592

    0.143 VA - 0.174 VB = 1.455

    ----------- ( 1 ) ----------- ( 2 )

    ----------- ( 3 ) 0.178 VA - 0.344 VC = 5.592

    VA , VB , VC

    -74-

  • Example 3 :- Solve the following circuit using nodal voltage method:

    9 50

    5

    30

    15

    20

    6

    15V

    3

    A C

    D

    B

    E

    Solution:

    Let D reference

    VA = 15 V

    B : 0201

    301

    615

    201

    301

    91

    61 =

    +

    +

    +

    +++ ECB VVV

    C : 0501

    315

    301

    151

    501

    31

    301 =

    +

    +

    +

    +++ EBC VVV

    D : 0501

    201

    51

    501

    201 =

    +

    +

    ++ CBE VVV

    Then rearrange the above equations and find VB , VC , VE .

    -75-

  • Example, (Sheet 4 Q. 24): For the following circuit diagram, find I & I1,

    using nodal voltage method:

    15V

    25

    15

    35

    17V

    15

    25

    12V

    0.8A

    0.3A

    20 A

    C

    D

    B

    DD

    A I1

    IB

    Solution:

    First; let D reference:-

    A : 03.02515

    4012

    4025201

    251

    401 =+++++

    ++ CBA VVV

    B : 02515

    25351

    151

    251 =+

    ++ IVV AB

    C : 08.04040

    12401 =+++

    IVV AC

    VB + 17 = VC

    Then the above equations can be minimized to:-

    B + C:

    ( ) 0401178.0

    4012

    401

    251

    2515

    351

    151

    251 =

    ++

    ++

    ++ BAB VVV

    ----------- ( 1 )

    ----------- ( 2 )

    ----------- ( 3 )

    ----------- ( 4 )

    ----------- ( 1 )

    -76-

  • A :

    ( )03.0

    2515

    4012

    4017

    25201

    251

    401 =++++++

    ++ BBA VVV

    Then solve to find VA & VB; hence we can find VC from eq. ( 4 ); to find I sub.

    VA & VB in eq. ( 2 ); and to find I1; then I1 = I + 0.8

    Second solution; let B reference

    Hence VC = 17 V

    A : 03.02515

    204017

    4012

    201

    251

    401 =+++++

    ++ DA VV

    D : 08.03.02020

    1351

    151 =+

    ++ AD VV

    Then find VA & VD from the above equations; hence to find I;

    C : 040128.0

    404017 =+++ IVA

    And to find I1; I1 = I + 0.8

    ----------- ( 2 )

    ----------- ( 1 )

    ----------- ( 2 )

    -77-

  • )9:( . .

    /

    Network Theorems:-

    1- Superposition Theorem:-

    In any circuit network contain more than one sources ( voltage or

    current ) to find the current ( or voltage ) in a certain part of a network , remove

    the sources of the network and find the current ( or voltage ) in the existence of

    only one source each time. The resultant current ( or voltage ) will be the

    algebraic sum of current ( or voltage ) due to all sources when acting

    independently once a time .

    (Removing the sources means:- Short circuiting the voltage source and open

    circuiting the current source) .

    Example 1:- In the following circuit diagram, find all branch current's using

    superposition theorem:-

    3

    6

    7

    25V

    3A4

    I5I2

    I1I4

    I3

    -79-

  • Solution :-

    1.) Effect of 25 V source :-

    AI 5.237

    251 =+=

    AI 5.264

    252 =+=

    AIII 5213 =+=

    2.) Effect of 3 A source :-

    3

    6

    7

    3A4

    I"2

    I"1I"4

    I"3

    I"5

    AI 1.237

    7*31 =+=

    AI 9.037

    3*34 =+=

    AI 2.164

    4*32 =+=

    AI 8.164

    6*35 =+=

    AIIIII 9.02.11.245213 ====

    3.) Superpose :-

    AIII 6.41.25.2111 =+=+= AIII 3.15.22.1222 === AIII 3.48.15.2525 =+=+= AIII 6.19.05.2414 === AIII 9.59.05333 =+=+=

    -80-

  • Example 2:- For the following circuit network, find the current in all branches,

    using superposition theorem:-

    30

    6040

    150V 270V

    I1

    I2

    I3

    Solution:-

    1.) Effect of 150 V source:-

    30

    6040

    150V

    I'1

    I'2

    I'3

    =++= 60306030*6040TR

    AI 5.260

    1501 ==

    AI 83.03060

    30*5.22 =+=

    AI 67.13060

    60*5.23 =+=

    -81-

  • 2.) Effect of 270 V source:-

    =++= 54604060*4030TR

    AI 554270

    1 ==

    AI 26040

    40*52 =+=

    AI 36040

    60*53 =+=

    3.) Superpose :-

    AIII 5.035.2311 === AIII 83.2283.0222 =+=+= AIII 33.3567.1133 === Example 3:- Find the current in all branch in the following circuit diagram:-

    4

    15V

    8

    5

    2A

    12V

    I4

    I2

    I3

    I1

    I5

    -82-

  • Solution:-

    1.) Effect of 15 V source:-

    AI 12.2

    858*54

    151 =

    ++=

    AI 3.185

    8*12.22 =+=

    AI 82.085

    5*12.23 =+=

    2.) Effect of 12 V source:-

    8

    5

    12V

    I"2

    I"1

    I"3

    4

    AI 57.1

    484*85

    121 =

    ++=

    AI 52.048

    4*57.12 =+=

    AI 04.148

    8*57.13 =+=

    -83-

  • 3.) Effect of 2 A source:-

    48

    5

    2A

    I'"3

    I'"2

    I'"5

    I'"4

    I'"1

    Vo

    =++= 74.151

    41

    811

    TT

    RR

    VRV To 5.3*2 ==

    AVI o 88.041==

    AVI o 7.052==

    AVI o 44.083==

    AII 12.12 14 == AII 3.12 25 ==

    3.) Superpose :-

    AIIII 04.012.104.112.24311 === AIIII 97.07.057.13.12122 === AIIII 03.13.157.13.15123 =+=+= AIIII 9.044.052.082.03234 =+=+= AIIII 96.188.004.112.21315 =+=+=

    -84-

  • )10:( . .

    /

    2-) Thevenin's Theorems:-

    .

    Any two terminal linear network can be replaced by an equivalent circuit

    of a voltage source ( Eth ) and a series resistor ( Rth ); as shown in figure below:-

    Hence; Lth

    th

    RREI +=

    Steps to find Eth & Rth :-

    1. Remove that portion of the network across which the Thevenins

    equivalent circuit is to be find.

    2. Mark the terminals of the remaining two terminal network.

    3. Calculate Rth by first setting all sources to zero ( voltage sources are

    replaced by short circuits and current sources are replaced by open

    circuit ), and finding the resultant resistance between the two marked

    terminals.

    4. Calculate Eth by first returning all sources to their origin positions and

    finding the open circuit voltage between the marked terminals.

    5. Draw the Thevenins equivalent circuit with the portion of the circuit

    previously removed replaced between the terminals of the equivalent

    circuit.

    -85-

  • Example 1:- For the following circuit diagram, find the current in ( 6 )

    resistor?

    54

    2

    25V

    63

    7A

    Solution:-

    54

    2

    25V

    3

    7A

    A B

    1.) Find Rth :

    54

    2

    3 A B

    Rth

    ( ){ }

    =+=+

    +=

    ++=

    22.759205

    454*5

    54//32thR

    -86-

  • 2.) Find Eth :

    A925

    ( )

    VE

    VV

    V

    VVV

    th

    oc

    oc

    oc

    89.23

    89.23359

    100

    05*7925*4

    021

    ===

    =+

    =+

    6

    Rth = 7.22

    Eth = 23.89 V

    A

    B

    I

    A

    RREI

    Lth

    th

    8.1622.7

    89.23 =+=

    +=

    -87-

  • Example 2:- Find the current in the 25 resistor for the following circuit

    network?

    10 40

    25

    20 20

    2V

    Solution:-

    1.) Find Rth :

    10 40

    20 20

    A B

    ( ) ( )

    =+++=

    +=

    20204020*40

    201020*10

    20//4020//10

    th

    th

    R

    R

    2.) Find Eth :

    10 40

    20 20

    A B2V

    Voc

    A602

    A302

    VE

    VV

    V

    th

    oc

    oc

    67.0

    67.06040

    3020

    6080

    0602*40

    302*10

    =

    ===

    =

    +

    -88-

  • AI

    RREI

    Lth

    th

    4567.0

    252067.0 =+=

    +=

    Example 3:- Find I in the ( 9 ) resistor for the following cct. diagram?

    9

    8

    7

    6A

    4A

    10

    25V

    Solution :-

    =++= 258107thR

    -89-

  • 8

    7

    6A

    4A

    10

    25V

    A B

    Voc

    6A

    4A

    10A

    ( ) ( ) ( )VV

    V

    oc

    oc

    201

    06*82510*107*4

    =

    =

    A

    RREI

    Lth

    th

    91.5925

    201 =+=

    +=

    -90-

  • Norton's Theorems:-

    Any two terminal linear network can be replaced by an equivalent circuit

    consisting of a current source and a parallel resistor.

    ththRE

    RN = Rth as before .

    IN = Isc = short circuit current between the two terminals of the active network.

    Example 1:- Find the current in 25 resistor for the following circuit network

    using Norton's Theorem?

    10 40

    25

    20 20

    2V

    -91-

  • Solution:-

    First find RN :-

    10 40

    20 20

    A B

    Second find IN :-

    10 40

    20 20

    A B2VIN

    I1 I2

    I4I3

    I

    )( ) (

    =+++=

    +=

    20204020*40

    201020*10

    20//4020//10NR

    AI91

    1082

    202020*20

    401040*10

    2 =+=+++

    =

    4020*

    91&

    4020*

    91

    5010*

    91&

    5040*

    91

    43

    21

    ==

    ==

    II

    II

    KC

    I1 IN I3 = 0

    L at A

    IN = I1 I3

    A033.04020*

    91

    5040*

    91 =

    =

    AIL 0147.0252020*033.0 =+=

    -92-

  • Example 2:- Find I in 50v voltage source, for the following circuit using

    eorem? Norton's Th

    25

    12

    20

    17

    30

    65V 50V45V

    Solution:-

    1.) Find RN :-

    12

    20

    17

    30

    25

    A

    B

    R1 R2 R3

    30 20A

    B

    -93-

  • == 8.754

    25*171R , == 78.354

    17*122R , == 56.554

    12*253R

    2.) Find IN :-

    ( ) ( )[ ][ ] =+=

    +++=

    1978.356.25//8.37

    20//30 231 RRRRN

    12

    20

    17

    30

    65V

    25

    45V

    A

    BIN

    Ib

    Ic

    Ia

    -47Ia + 17Ic + 65 = 0

    -32Ib + 12Ic - 45 = 0

    -54Ic + 17Ia + 12Ib = 0

    After find Ia , Ib , Ic

    IN = Ia Ib

    195050 I -I I

    0 I-I-I

    aNL

    LaN

    ====

    NN

    N IRI

    -94-

  • Maximum Power Transfer:-

    A load will receive maximum power from a d.c. network when its total

    resistive value is exactly equal to the Thevenin resistance of the network.

    For Thevenin cct. Nortan cct.

    For Max. power

    LLth

    thLLL RRR

    ERIP *2

    2.max

    +==

    thth

    th RRE *

    4 22

    =

    RNIN RL

    LLN

    NNLLL RRR

    RIRIP *2

    2.max

    +==

    NN

    NN RREI *

    4 22

    2=

    RL = Rth RL = RN

    Under Max. Power transfer conditions, the efficiency is:-

    %100*i

    o

    PP % =

    %100*Lth

    LL

    IEIV= %100*

    th

    L

    EV=

    th

    thL R

    EP4

    2

    .max=

    4

    2

    .max

    NNL

    RIP =

    -95-

  • Eth = VL + Rth Ith

    max. power transfer )

    th = VL + RL IL

    = VL + VL = 2VL

    Q Rth = RL ( for

    E

    %50%100100*=thE

    *2

    % ==L

    LL

    VVV

    h efficiency will always be 50% under max. power transfer conditions .

    Practical example:-

    T e *

    RLEth

    RthI

    et Rth = 3 & RL = 1 & Eth = 15 V L

    WPL 141*1315 2

    +=

    For RL = 2 WPL 182*515 2 =

    =

    For RL = 3 WPL 75.183*615 2 =

    =

    LL RIP

    2=

    LLth

    RRR

    E *

    +=

    2

    -96-

  • For RL = 4 WP 36.184*15 2 == L 7

    WPL 57.175*815 2 =

    =For RL = 5

    L = Rth , we get the max. power of PL .

    ence

    Note that when R

    H EIPP inL

    1122

    ==

    r Pin = 2 PL Example 1:- Find the value of RL for maximum power transfer to RL , and

    determine the power delivered under these conditions ?

    o

    Solution:-

    L , and find the equivalent resistance ( Rth ) First remove R

    -97-

  • (

    For Max. power RL = Rth

    = 14LR 12

    6

    3

    18V

    Vab

    a

    b

    Find the value of R for the following cct. for max. power transfer,

    L

    Example 2:- L and find P ?

    Solution:-

    8

    7

    4

    6Rth

    a

    b

    )

    =+ +=

    +

    141263

    12

    th

    R = 6//3th

    6*3R

    VVE abth 12366*18 =+==

    ( ) WREP

    th

    th 57.214*4

    124

    22

    .max ===

    -98-

  • ( ) ( )

    V

    EVV thaboc

    7215*25

    120

    78*7846

    120

    =

    ++++===

    =

    ( ) WRIP LL 2166*6672

    22 =

    +==

    or

    RL= 6

    Eth = 72 V

    Rth = 6

    a

    b

    ( ) ( )

    L

    theq

    R

    RR

    ==+=

    +==

    6101510*15

    46//7//8.

    WREP

    th

    thL 2166*4

    74

    22

    ===

    -99-

  • Example 3 (sheet 5, fig. 20):- Find the maximum power in ( R ), for the

    following cct. diagram?

    1

    1 2

    6 6

    2A4

    10V

    6V

    2

    2 4

    R

    6V

    Solution:-

    1-) Find Rth.:-

    -100-

  • RRR theq ==+++== 102314. 2.) Find Eth :-

    -101-

  • 1

    1 2

    6 6

    2A4

    10V

    6V

    2

    2 4

    6V

    A

    B

    IY

    IX

    IZ

    2A

    Voc

    ( )( )

    ( ) AIIAII

    AII

    zz

    yy

    xx

    25.102*268

    33.106*261012

    202*1104

    ==++

    ==++

    ==+

    From KVL ( ) ( ) ( )

    VV

    V

    oc

    oc

    154856

    02*233.1*625.1*46

    =++=

    =++

    AI 75.01010

    15 =+=

    -102-

  • Rth = 10 RL= 10

    Eth = 15 V

    B

    AI

    Example 4 (sheet 5, fig. 21):- Find the maximum power in ( R ), for the

    following cct. diagram?

    ( ) WRIP

    625.510*75.0 2

    2.max

    ==

    =

    80mA18

    5

    12

    R

    15

    0.86V

    0.3V

    8

    20mA

    Solution:-

    First find Req.:- // )515( +)1812(. +=eqR //8 = 4.8

    -103-

  • Second find IN:-

    18

    5

    12

    15

    0.86V

    0.3V

    0.36V

    1.2V

    8

    Isc A

    B

    30

    20

    0.5V

    1.5V

    A

    B

    Isc

    IY

    IX

    mAII xysc 205.1 ==II

    II

    II

    N

    yy

    xx

    3.5830

    5.0

    205.105.120

    305.005.030

    ==

    ==+

    ==+

    -104-

  • A2

    3.58

    mWP 1.48.4*10*2

    3.58 23.max =

    =

    -105-